A car with good tread can stop in less distance than a car with poor tread. A model for the stopping distance d, in feet, of a car with good tread on dry cement is

d = 0.04v2 + 0.5v,
where v is the speed of the car in miles per hour. If the driver must be able to stop within 65 ft, what is the maximum safe speed of the car? Round to the nearest whole number.

d=65ft=.04v^2+.05v

put this into quadratic form, and use the quadratic equation, solve for v.

To find the maximum safe speed of the car, we need to solve the equation for speed when the stopping distance is equal to 65 ft.

The given model for stopping distance is: d = 0.04v^2 + 0.5v

Setting d = 65, we have:

65 = 0.04v^2 + 0.5v

To solve this quadratic equation, we can rearrange it to the standard quadratic form:

0.04v^2 + 0.5v - 65 = 0

Now, we can either factorize or use the quadratic formula. In this case, let's use the quadratic formula:

v = (-b ± √(b^2 - 4ac)) / 2a

For our equation, a = 0.04, b = 0.5, and c = -65.

Plugging these values into the quadratic formula, we have:

v = (-0.5 ± √(0.5^2 - 4*0.04*(-65))) / (2*0.04)

Simplifying further:

v = (-0.5 ± √(0.25 + 10.4)) / 0.08
v = (-0.5 ± √10.65) / 0.08

Now, we can calculate the two possible values for v:

v₁ = (-0.5 + √10.65) / 0.08
v₂ = (-0.5 - √10.65) / 0.08

Using a calculator, we find:

v₁ ≈ 64.34 mph
v₂ ≈ -81.91 mph

Since speed cannot be negative in this context, we discard the negative value. Therefore, the maximum safe speed of the car is approximately 64 mph (rounded to the nearest whole number).